Implicit Differentiation to find dy/dx

In summary, to use implicit differentiation, treat all variables as functions of x and differentiate both sides of the equation with respect to x. Use the chain rule for any terms with y, and then solve for dy/dx.
  • #1
akstradtne
2
0
Use implicit differentiation to find dy/dx given x^2y+xy^2=4.

I have no idea how to approach this problem. My instructor assigned this as homework but has not gone over it at all in class. We have gone over explicit differentiation and I understand this well. I have read the section but it is making no sense. I think I need to use the General Power Rule: d/dx(y^n)=ny^(n-1)dy/dx but I don't know how to use it.
 
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  • #2
akstradtne said:
Use implicit differentiation to find dy/dx given x^2y+xy^2=4.

I have no idea how to approach this problem. My instructor assigned this as homework but has not gone over it at all in class. We have gone over explicit differentiation and I understand this well. I have read the section but it is making no sense. I think I need to use the General Power Rule: d/dx(y^n)=ny^(n-1)dy/dx but I don't know how to use it.

You've got all the ingredients. Differentiate both sides with respect to x and solve for dy/dx. Use the product rule on the way.
 
  • #3
Can you explain a little more. I don't understand "with respect to x or y." How do I use the general power rule?
 
  • #4
You said, with reference to implicit differentiation, "I understand this well". Did you mean to say "I don't understand this well"?

Differentiating "with respect to x" means treating y as a function of x, not a separate variable. So "[itex]y^n[/itex], differentiated "with respect to y" would be just [itex]d(y^n)/dy= ny^{n-1}[/itex]. But differentiating "with respect to x" you would use the chain rule: [tex]d(y^n)/dx= ny^{n-1} (dy/dx)[/tex].

If you had, for example, [itex]x^2y+ xy^2= x- y[/itex], differentiating each part with respect to x, (using the "product rule" for the terms on the left), [itex]2xy+ x^2(dy/dx)+ (1)y^2+ 2xy(dy/dx)= 1- dy/dx[/itex]. You can then solve that equation for dy/dx.
 

FAQ: Implicit Differentiation to find dy/dx

What is implicit differentiation and how is it used to find dy/dx?

Implicit differentiation is a method used in calculus to find the derivative of a function that is not explicitly expressed in terms of x. It involves treating each variable as a function and using the chain rule to find the derivative with respect to x. The resulting derivative is dy/dx, the rate of change of y with respect to x.

2. Can implicit differentiation be used for any type of function?

Yes, implicit differentiation can be used for any type of function, as long as it is differentiable. This means that the function must be continuous and have a defined derivative at every point.

3. How do you perform implicit differentiation?

To perform implicit differentiation, you must first identify the variables in the function and treat each one as a function. Then, use the chain rule to find the derivative of each variable with respect to x. Finally, solve for dy/dx by rearranging the resulting equation.

4. When is implicit differentiation necessary?

Implicit differentiation is necessary when the dependent variable, y, cannot be easily isolated on one side of the equation. This often occurs when the function is expressed as an equation with both x and y terms, such as in implicit equations or implicit functions.

5. Are there any common mistakes to avoid when using implicit differentiation?

Yes, there are a few common mistakes to avoid when using implicit differentiation. These include not properly applying the chain rule, forgetting to include the derivative of y with respect to x (dy/dx), and making algebraic errors when solving for dy/dx. It is important to carefully follow each step and check your work to avoid these mistakes.

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